Differentiate the following w.r.t.x
sin 3x (ii) `cos^2x` (iii) `sin x^3`

Answer

(i) `3cos 3x` (ii) - sin 2 x (iii) `3x^2 cos x^3`

Explanation

(i) `y = sin 3 x,` Let `u = 3x , (du)/(dx) = 3`
Then, `y = sin u, (dy)/(du) = cos u = cos 3x :. (dy)/(dx) = (dy)/(du)xx(du)/(dx) =(cos 3x)xx3 = 3 cos 3x`
(ii) Let `u = cos x,(du)/(dx) = - sin x, y= cos^2 x = u^2 , (dy)/(du) =2u = 2 cos x`
`(dy)/(dx) = (dy)/(du)xx(du)/(dx) =2cos x (-sin x)=-sin 2x`
(iii) Let `u =x^3 , (du)/(dx)=3x^2, y= sinu, (dy)/(dx) = cos u =cos x^3`
`(dy)/(dx) = (dy)/(du)xx(du)/(dx) = 3x^2 cos x^3`